Adama Science &Technology University

Department of Chemical Engineering

FLUID MECHANICS FOR CHEMICAL ENGINEERS

(ChE 2206)

May 2022
© EA

22/07/2022 By EA 1
5. Dimensional Analysis
• Upon completion of this chapter you will
be able to
–Explain dimensional analysis
–Differentiate different methods of
dimensional analysis
–Explain similitude

22/07/2022 By EA 2
 Introduction
• Dimensional analysis is a mathematical
technique making use of study of dimensions.
• is used in research work for design and for
conducting model tests.
• deals with the dimensions of physical
quantities involved in the phenomenon.
• predicts the physical parameters that will
influence the flow,
• indicates the direction in which the
experimental work should go.
Types of Dimensions
• There are two types of dimensions
– Fundamental Dimensions/Quantities
– Secondary Dimensions or Derived Quantities
• Fundamental Dimensions/ Quantities: These
are basic quantities.
• For Example;
– Time,T
– Distance, L
– Mass, M
Types of Dimensions
Secondary Dimensions or Derived Quantities
• possess more than one fundamental dimension.
• For example;
– Velocity is denoted by distance per unit time L/T
– Acceleration is denoted by distance per unit time
square L/T2
– Density is denoted by mass per unit volume M/L3
• Since velocity, density and acceleration involve
more than one fundamental quantities so these
are called derived quantities.
Methodology of Dimensional Analysis
Dimensionally Homogeneous equation
• The dimensions of each terms in an equation on
both sides are equal.
• are independent of system of units.
• For example; Let’s consider the equation V=(2gH)1/2
–Dimensions of LHS=V=L/T=LT-1
–Dimensions of RHS=(2gH)1/2=(L/T2xL)1/2=LT-1
–Dimensions of LHS= Dimensions of RHS
•So the equation V=(2gH)1/2 is dimensionally
homogeneous equation.
 Methods of Dimensional Analysis
•The relation among the variables can be determined
by the following two methods;
–Rayleigh’s Method
–Buckingham’s π-Theorem
Rayleigh’s Method:
•It is used for determining expression for a variable
(dependent) which depends upon maximum three to
four variables (Independent) only.
•If the number of independent variables are more than
4 then it is very difficult to obtain expression for
dependent variable.
 Methods of Dimensional Analysis
•Let X is a dependent variable which depends upon
X1, X2, and X3 as independent variables. Then
according to Rayleigh’s Method

X=f(X1, X2, X3) which can be written as

X=K X1a, X2b, X3c

Where K is a constant and a, b, c are arbitrary powers

which are obtained by comparing the powers of
fundamental dimensions.
22/07/2022 By EA 8
 Rayleigh’s Method
The resisting force R of a supersonic plane during flight can be
considered as dependent upon the length of the aircraft l, velocity
V, air viscosity μ, air density ρ, and bulk modulus of air k. Express
the functional relationship between the variables and the resisting
force.
◼ Solution:
R = f (l ,V ,  ,  , K )  R = Al a , V b ,  c ,  d , K e (1)
Where: A = Non dimensional constant
Substituting the powers on both sides of the equation
MLT -2 = ALa ( LT −1 )b ( ML−1T −1 )c ( ML−3 ) d ( ML−1T −2 )e
Equating the powers of MLT on both sides
Power of M  1 = c + d + e
Power of L  1 = a + b - c - 3d - e
Power of T  − 2 = -b - c - 2e
Rayleigh’s Method
Since the unkown(5) are more than number of equations(3). So expressing
a, b & c in terms of d & e
d = 1- c - e
b = 2 - c - 2e
a = 1- b + c + 3d + e = 1- (2 - c - 2e) + c + 3(1- c - e) + e
= 1- 2 + c + 2e + c + 3 - 3c - 3e + e = 2 - c
Substituting the values in (1), we get
R = Al 2−cV 2−c − 2 e  c  1−c −e K e = Al 2V 2  (l − cV − c  c  − c )(V −2e  − e K e )
 c
2 2     K 
e

R = A l V    2 

 Vl   V  

2 2     K  
R = A l V    2 
 Vl  V 
 Buckingham’s π-Theorem:
• Rayleigh’s Method becomes laborious if variables are more
than fundamental dimensions (MLT),
• The difficulty is overcome by Buckingham’s π-Theorem
which states that
• “If there are n variables (Independent and Dependent) in a
physical phenomenon and if these variables contain m
fundamental dimensions then the variables are arranged
into (n-m) dimensionless terms which are called π-terms.”
• Let X1, X2, X3,…,X4, Xn are the variables involved in a
physical problem.
Buckingham’s π-Theorem:
• Let X1 be the dependent variable and X2, X3, X4,…,Xn
are the independent variables on which X1 depends.
Mathematically it can be written as
X1=f(X2 ,X3 ,X4 ,Xn) which can be rewritten as
f1(X1,X2 X3 X4 Xn)=0
• Above equation is dimensionally homogenous.
• It contain n variables and if there are m fundamental
dimensions then it can be written in terms of dimensions
groups called π-terms which are equal to (n-m)
• Hence f1(π1 π2 π3,… πn-m)=0

22/07/2022 By EA 12
Buckingham’s π-Theorem:
• Properties of π-terms:
–Each π-term is dimensionless and is independent of system of units.
–Division or multiplication by a constant does not change the character of the π-
terms.
–Each π-term contains m+1 variables, where m is the number of fundamental
dimensions and also called repeating variable.
•Let in the above case X2,X3,X4 are repeating variables and if fundamental dimensions
m=3 then each π-term is written as
Π1=X2a1.X3b1.X4a1 .X1
Π2=X2a2.X3b2.X4a2 .X5
.
.
Πn-m=X2a(n-m).X3b(n-m).X4a(n-m) .Xn
Each equation is solved by principle of dimensionless homogeneity and values of a1, b1
& c1 etc are obtained. Final result is in the form of
Π1=(Π2, Π3, Π4 ,…, Π(n-m))
Π2=(Π1, Π3, Π4 ,…, Π(n-m))
Methods of Selecting Repeating
Variables
• The number of repeating variables are equal to number of
fundamental dimensions of the problem. The choice of repeating
variables is governed by following considerations;
– As far as possible, dependent variable shouldn’t be selected as
repeating variable.
– The repeating variables should be chosen in such a way that one
variable contains geometric property, other contains flow
property and third contains fluid property.
– The repeating variables selected should form a dimensionless
group
– The repeating variables together must have the same number of
fundamental dimension.
– No two repeating variables should have the same dimensions.
• Note: In most of fluid mechanics problems, the choice of repeating
variables may be (i) d,v ρ, (ii) l,v,ρ or (iii) d, v, μ.
Buckingham’s π-Theorem:
The resisting force R of a supersonic plane during flight can be considered as dependent upon the length of

the aircraft l, velocity V, air viscosity μ, air density ρ, and bulk modulus of air k. Express the functional

relationship between the variables and the resisting force.

R = f (l , V ,  ,  , K )  f ( R, l , V ,  ,  , K ) = 0
Total number of variables, n= 6
No. of fundamental dimension, m=3
No. of dimensionless  -terms, n-m=3
Thus: f ( 1 ,  2 ,  3 ) = 0
No. Repeating variables =m=3
Repeating variables =l , V , 
Thus π-terms are written as
 1 = l a1V b1  c1 R
 2 = l a 2V b 2  c 2 
 3 = l a 3V b 3  c 3 K
Buckingham’s π-Theorem:
• Now each Pi-term is solved by the principle of dimensional
homogeneity
 1 − term  M o LoT o = La1 ( LT −1 )b1 ( ML−3 ) c1 MLT −2
Equating the powers of MLT on both sides, we get
Power of M: 0=c 1 +1  c 1 =-1
Power of L: 0=a 1 +b1 -3c1 +1  a1 = −2
Power of T: 0=-b1 -2  b1 =-2
R
  1 = l -2V -2
 -2 R   1 =
 L2V 2
 2 − term  M o LoT o = La 2 ( LT −1 )b 2 ( ML−3 ) c 2 ML−1T −1
Equating the powers of MLT on both sides, we get
Power of M: 0 = c 2 +1  c 2 = -1
Power of L: 0 = a2 + b2 - 3c2 -1  a2 = −1
Power of T: 0 = -b2 -1  b2 = -1

  2 = l -1V -1
 -1   2 =
 lV
Buckingham’s π-Theorem:
 3 − term  M o LoT o = La 3 ( LT −1 )b 3 ( ML−3 )c 3 ML−1T −2
Equating the powers of MLT on both sides, we get
Power of M: 0 = c 3 +1  c 3 = -1
Power of L: 0 = a3 + b3 - 3c3 -1  a3 = −0
Power of T: 0 = -b3 - 2  b3 = -2
K
  3 = l 0V -2  -1 K   2 =
V 2
Hence
 R  K 
f ( 1 2 3 ) = f  2 2 , , 2  = 0 or
  l V lV  V  
R   K    K 
=  , 2   R = l V  
2 2
, 2 
l V
2 2
 lV  V    lV  V  
 Similitude and Model Analysis
• Similitude is a concept used in testing of Engineering Models.
• Usually, it is impossible to obtain a pure theoretical solution of
hydraulic phenomenon.
• Therefore experimental investigations are often performed on small
scale models, called model analysis.
• A few examples, where models may be used are
– ships in towing basins,
– air planes in wind tunnel,
– hydraulic turbines,
– centrifugal pumps,
– spillways of dams,
– river channels etc
• to study such phenomenon as the action of waves and tides on
beaches, soil erosion, and transportation of sediment etc.
Model Analysis
• Model: is a small scale replica of the actual structure.
• Prototype: the actual structure or machine.
• Note: It is not necessary that the models should be smaller
that the prototype, they may be larger than prototype.

Lp1
Lm1

Lp2 Lm2
Fp1 Fp2 Fm1 Fm2

Fm3 Lm3
Fp3 Lp3
Prototype Model
Similitude-Type of Similarities
• Similitude: is defined as similarity between the
model and prototype in every respect, which mean
model and prototype have similar properties or
model and prototype are completely similar.
• Three types of similarities must exist between
model and prototype.
– Geometric Similarity
– Kinematic Similarity
– Dynamic Similarity
Similitude-Type of Similarities
• Geometric Similarity: is the similarity of shape. It is said to exist
between model and prototype if ratio of all the corresponding linear
dimensions in the model and prototype are equal. E.g.
Lp Bp Dp
= = = Lr
Lm Bm Dm

◼ Where: Lp, Bp and Dp are Length, Breadth, and diameter of prototype and Lm,
Bm, Dm are Length, Breadth, and diameter of model.
◼ Lr= Scale ratio

◼ Note: Models are generally prepared with same scale ratios in every
direction. Such a model is called true model. However, sometimes it is not
possible to do so and different convenient scales are used in different
directions. Such a models is call distorted model
Similitude-Type of Similarities
• Kinematic Similarity: is the similarity of motion. It is said to exist
between model and prototype if ratio of velocities and acceleration at
the corresponding points in the model and prototype are equal. E.g.

V p1 Vp 2 a p1 ap2
= = Vr ; = = ar
Vm1 Vm 2 am1 am 2
◼ Where: Vp1& Vp2 and ap1 & ap2 are velocity and accelerations at point 1 & 2
in prototype and Vm1& Vm2 and am1 & am2 are velocity and accelerations at
point 1 & 2 in model.
◼ Vr and ar are the velocity ratio and acceleration ratio

◼ Note: Since velocity and acceleration are vector quantities, hence not only
the ratio of magnitude of velocity and acceleration at the corresponding
points in model and prototype should be same; but the direction of
velocity and acceleration at the corresponding points in model and
prototype should also be parallel.
Similitude-Type of Similarities
• Dynamic Similarity: is the similarity of forces. It is said to exist
between model and prototype if ratio of forces at the
corresponding points in the model and prototype are equal. E.g.
( Fi ) p ( Fv ) p ( Fg ) p
= = = Fr
( Fi )m ( Fv )m ( Fg )m
◼ Where: (Fi)p, (Fv)p and (Fg)p are inertia, viscous and gravitational forces in
prototype and (Fi)m, (Fv)m and (Fg)m are inertia, viscous and gravitational forces in
model.
◼ Fr is the Force ratio

◼ Note: The direction of forces at the corresponding points in model and

prototype should also be parallel.
Types of forces encountered in
fluid Phenomenon
•Inertia Force, Fi: It is equal to product of mass and
acceleration in the flowing fluid.
•Viscous Force, Fv: It is equal to the product of shear stress
due to viscosity and surface area of flow.
•Gravity Force, Fg: It is equal to product of mass and
acceleration due to gravity.
•Pressure Force, Fp: it is equal to product of pressure
intensity and cross-sectional area of flowing fluid.
•Surface Tension Force, Fs: It is equal to product of surface
tension and length of surface of flowing fluid.
•Elastic Force, Fe: It is equal to product of elastic stress and
area of flowing fluid.
Dimensionless Numbers
• These are numbers which are obtained by dividing the
inertia force by viscous force or gravity force or pressure
force or surface tension force or elastic force.
• As this is ratio of once force to other, it will be a
dimensionless number. These are also called non-
dimensional parameters.
• The following are most important dimensionless numbers.
– Reynold’s Number
– Froude’s Number
– Euler’s Number
– Weber’s Number
– Mach’s Number
Dimensionless Numbers
• Reynold’s Number, Re: It is the ratio of inertia force
to the viscous force of flowing fluid.
Velocity Volume
Fi
Mass.  . Velocity
Re = = Time = Time
Fv Shear Stress. Area Shear Stress. Area
 QV.  AV .V  AV .V VL VL
= = = = =
 .A du
 .A  .A
V  
dy L

◼ Froude’s Number, Fe: It is the ratio of inertia force to the

gravity force of flowing fluid.
Velocity Volume
Fi
Mass.  . Velocity
Fe = = Time = Time
Fg Mass. Gavitational Acceleraion Mass. Gavitational Acceleraion
 QV
.  AV .V V2 V
= = = =
Volume.g  AL.g gL gL
Dimensionless Numbers
• Eulers’s Number, Eu: It is the ratio of inertia force to the
pressure force of flowing fluid.
Velocity Volume
Fi
Mass.  . Velocity
Eu = = Time = Time
Fp Pr essure. Area Pr essure. Area
 QV
.  AV .V V2 V
= = = =
P. A P. A P/ P/

◼ Weber’s Number, We: It is the ratio of inertia force to the surface tension
force of flowing fluid.
Velocity Volume
Fi
Mass.  . Velocity
We = = Time = Time
Fg Surface Tensionper. Length Surface Tensionper. Length
 QV
.  AV .V  L2V 2 V
= = = =
 .L  .L  .L 
L
Dimensionless Numbers
• Mach’s Number, Re: It is the ratio of inertia force
to the elastic force of flowing fluid.

Velocity Volume
Fi
Mass.  . Velocity
M= = Time = Time
Fe Elastic Stress. Area Elastic Stress. Area
 QV
.  AV .V  L2V 2 V V
= = = = =
K .A K .A KL2 K/ C
Where : C = K / 
Model Laws or similarity Laws
• We have already read that for dynamic similarity ratio of corresponding
forces acting on prototype and model should be equal. i.e
( Fv ) p ( Fg ) p ( Fp ) p ( Fs ) p ( Fe ) p ( FI ) p
= = = = =
( Fv )m ( Fg )m ( Fp )m ( Fs )m ( Fe )m ( FI )m
◼ Force of inertial comes in play when sum of all other forces is not equal to
zero which mean

(F + F
v g + Fp + Fs + Fe ) = ( FI )
Thus dynamic similarity require that
(F + F
v g + Fp + Fs + Fe ) ( FI ) p
= p

(F + F
v g + Fp + Fs + Fe ) m ( FI )m
◼ In case all the forces are equally important, the above two equations cannot
be satisfied for model analysis
Model Laws or similarity Laws
•For practical problem only the most significant force is
considered for dynamic similarity.
•models are designed on the basis of ratio of force, which is
dominating in the phenomenon.
•Finally the laws on which models are designed for dynamic
similarity are called models laws or laws of similarity.
•The followings are these laws
–Reynold’s Model Law
–Froude’s Model Law
–Euler’s Model Law
–Weber’s Model Law
–mach’s Model Law
Reynold’s Model Law
• It is based on Reynold’s number and states that Reynold’s number for
model must be equal to the Reynolds number for prototype.
• Reynolds Model Law is used in problems where viscous forces are
dominant.These problems include:
– Pipe Flow
– Resistance experienced by submarines, airplanes, fully immersed bodies etc.
VP LP Vm Lm
( Re ) P = ( Re )m or =
P m
VP LP VL
= r r =1
 P  r
Vm Lm  

 m
VP LP P
where : Vr = , Lr = ,r =
Vm Lm m
Reynold’s Model Law
• The Various Ratios for Reynolds’s Law are obtained as
 VL   VL 
sin ce   =   and  =  / 
   P   m
V L  
Velocity Ratio: Vr = P = m P = r
Vm LP m L r
TP L /V L
Time Ratio: Tr= = P P = r
Tm L m /Vm Vr
aP VP / TP Vr
Acceleration Ratio: a r = = =
am Vm / Tm Tr
APVP
Discharge Ratio: Q r = = L2rVr
AmVm
Force Ratio: Fr =mr ar =  r QrVr =  r L2rVrVr =  r L2rVr2
Power Ratio: Pr =Fr .Vr = r L2rVr2Vr =  r L2rVr3
Reynold’s Model Law
A pipe of diameter 1.5 m is required to transport an oil of specific gravity 0.90
and viscosity 3x10-2 poise at the rate of 3000litre/sec. Tests were conducted on a
15 cm diameter pipe using water at 20oC. Find the velocity and rate of flow in
the model.
◼ Solution:
◼ Prototype Data:
For pipe flow,
◼ Diameter, Dp= 1.5m
According to Reynolds' Model Law
◼ Viscosity of fluid, μp= 3x10-2 poise
 m Vm D m  p Vp D p V D 
◼ Discharge, Qp =3000litre/sec =  m = p p m
m p Vp m Dm  p
◼ Sp. Gr., Sp=0.9
Vm 900 1.5 110−2
◼ Density of oil=ρp=0.9x1000 = = 3.0
V p 1000  0.15 3 10−2
=900kg/m3
Qp 3.0
◼ Model Data: Since Vp = =
Ap  / 4(1.5) 2
◼ Diameter, Dm=15cm =0.15 m
= 1.697 m / s
◼ Viscosity of water, μm =1x10-2 poise
Vm = 3.0V p = 5.091m / s
◼ Density of water, ρm=1000kg/m3n
◼ Velocity of flow Vm=? and Qm = Vm Am = 5.091  / 4(0.15) 2
= 0.0899m3 / s
◼ Discharge Qm=?
Reynold’s Model Law
A ship 300m long moves in sea water, whose density is 1030 kg/m3. A 1:100
model of this ship is to be tested in a wind tunnel. The velocity of air in the
wind tunnel around the model is 30m/s and the resistance of the model is
60N. Determine the velocity of ship in sea water and also the resistance of
ship in sea water. The density of air is given as 1.24kg/m3. Take the kinematic
viscosity of sea water and air as 0.012 stokes and 0.018 stokes respectively.
◼ Solution:
◼ For Prototype ◼ For Model
◼ Length, Lp= 300m ◼ Scale ratio = Lp/Lm=100
◼ Fluid = sea water ◼ Length, Lm= Lp/100 = 3m
◼ Density of sea water, ρp= 1030 kg/m3 ◼ Fluid = air
◼ Kinematic Viscosity, νp=0.018 stokes ◼ Density of air, ρm= 1.24 kg/m3
=0.018x10-4 m2/s ◼ Kinematic Viscosity, νm=0.012 stokes
◼ Let Velocity of ship, Vp =0.012x10-4 m2/s
◼ Resistance, Fp ◼ Velocity of ship, Vm=30 m/s
◼ Resistance, Fm = 60 N
Reynold’s Model Law
• For dynamic similarity between model and prototype, the
Reynolds number for both of them should be equal.
 VL   VL   p Lm
  =    Vp = Vm
   p   m m Lp
0.012 10−4 3
Vp = −4
30 = 0.2m / s
0.018 10 300
Since Resistance= Mass  Acceleration= L2 V 2
Fp ( L V )
2 2 2 2
1030  300   0.2 
Thus = p
=     = 369.17
Fm ( L V )
2 2
m
1.24  3   30 
Fp = 369.17  60 = 22150.2 N
Froude’s Model Law
• Froude’s number for model must be equal to the Froude’s number
for prototype.
• is used in problems where gravity forces is only dominant to control
flow in addition to inertia force.
• These problems include:
– Free surface flows such as flow over spillways, weirs, sluices, channels etc.
– Flow of jet from orifice or nozzle
– Waves on surface of fluid
– Motion of fluids with different viscosities over one another
VP Vm VP Vm
( F e )P = ( F e )m or = or =
g P LP g m Lm LP Lm
VP VP LP
= Vr / Lr = 1; where : Vr = , Lr =
 LP  Vm Lm
Vm  
 Lm 
Froude’s Model Law
• The Various Ratios for Froude’s Law are obtained as
VP V
sin ce = m
LP Lm
V Lp
Velocity Ratio: Vr = P = = Lr
Vm Lm
TP L /V L
Time Ratio: Tr= = P P = r = Lr
Tm L m /Vm Lr
aP V /T Vr Lr
Acceleration Ratio: a r = = P P = = =1
am Vm / Tm Tr Lr
APVP
Discharge Ratio: Q r = = L2rVr = L2r Lr = L5r / 2
AmVm
Force Ratio: Fr=mr ar =  r QrVr =  r L2rVrVr =  r L2rVr2 =  r L2r Lr =  r L3r

( )
3
Power Ratio: Pr=Fr.Vr= r L V Vr =  r L V =  r L
2
r r
2 2
r r
3 2
r Lr =  r L7r / 2
Froude’s Model Law
In the model test of a spillway the discharge and velocity of flow over
the model were 2 m3/s and 1.5 m/s respectively. Calculate the
velocity and discharge over the prototype which is 36 times the
model size.
◼ Solution: Given that For Discharge
Qp
= ( Lr ) = ( 36 )
2.5 2.5

Qm
Q p = ( 36 )  2 = 15552 m3 / sec
2.5

◼ For Model
◼ Discharge over model, Qm=2 m3/sec For Dynamic Similarity,
◼ Velocity over model, Vm = 1.5 m/sec Froude Model Law is used
◼ Linear Scale ratio, Lr =36 Vp
= Lr = 36 = 6
Vm
◼ For Prototype V p = 6 1.5 = 9 m / sec
◼ Discharge over prototype, Qp =?
◼ Velocity over prototype Vp=?
Numerical Problem:
The characteristics of the spillway are to be studied by means of a geometrically
similar model constructed to a scale of 1:10.
– (i) If 28.3 cumecs, is the maximum rate of flow in prototype, what will be the corresponding flow
in model?
– (i) If 2.4m/sec, 50mm and 3.5 Nm are values of velocity at a point on the spillway, height of
hydraulic jump and energy dissipated per second in model, what will be the corresponding velocity
height of hydraulic jump and energy dissipation per second in prototype?

◼ Solution: Given that

For Model
◼ Discharge over model, Qm=?
◼ Velocity over model, Vm = 2.4 m/sec
◼ Height of hydraulic jump, Hm =50 mm
◼ Energy dissipation per second, Em =3.5 Nm
◼ Linear Scale ratio, Lr =10
◼ For Prototype
◼ Discharge over model, Qp=28.3 m3/sec
◼ Velocity over model, Vp =?
◼ Height of hydraulic jump, Hp =? Energy dissipation per second, Ep =?
Froude’s Model Law
For Discharge: For Hydraulic Jump:
Qp Hp
= L2.5
r = 10
2.5
= Lr = 10
Qm Hm
Qm = 28.3 /102.5 = 0.0895 m3 / sec H p = 50 10 = 500 mm
For Velocity: For Energy Dissipation:
Vp Ep
= Lr = 10 = L3.5
r = 10 3.5
Vm Em
V p = 2.4  10 = 7.589 m / sec E p = 3.5 103.5 = 11067.9 Nm / sec
 The End!

22/07/2022 By EA 41